\newproblem{lay:6_4_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.4.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A=\begin{pmatrix}5&9\\1&7\\-3&-5\\1&5\end{pmatrix}$ and $Q=\begin{pmatrix}\frac{5}{6}&-\frac{1}{6}\\ \frac{1}{6}& \frac{5}{6}\\-\frac{3}{6}&\frac{1}{6}\\
		\frac{1}{6}& \frac{3}{6}\end{pmatrix}$. The columns of $Q$ were obtained by applying the Gram-Schmidt process to the columns of $A$. Find an upper triangular
		matrix $R$ such that $A=QR$.
}{
   % Solution
	Since $Q$ is an orthogonal matrix, its inverse is its transpose $Q^TQ=I$. Then, we simply multiply the decomposition $A=QR$ by $Q^T$ on the left to obtain
	\begin{center}
		$A=QR$ \\
		$Q^TA=R$ \\
		$R=\begin{pmatrix}\frac{5}{6} & \frac{1}{6} & -\frac{3}{6} & \frac{1}{6}\\ -\frac{1}{6} & \frac{5}{6} & \frac{1}{6} & \frac{3}{6}\end{pmatrix}
		   \begin{pmatrix}5&9\\1&7\\-3&-5\\1&5\end{pmatrix}=\begin{pmatrix}6&12\\0 & 6\end{pmatrix}$
	\end{center}
}
\useproblem{lay:6_4_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
